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The formula for gravitational potential energy, $$-G\frac{m_1 m_2}{R},$$ is found by using the fact that the change in potential energy is equal to negative of the work done ( by conservative forces). One of the assumptions is that the 2nd larger mass remains stationary relative to each other, and thus only the work done on the much smaller has to be taken into account.

This is obviously true for something like A satellite and the earth, but what about the case when the masses are similar?

Is this formula still when the two gravitating masses are of similar mass?

I tried to derive the formula without the assumption by adding a pseudo force on $m_2$ to take $m_1$ at rest. I arrive at the formula $$U(R) = -G \frac{m_2 (m_1 + m_2)}{R},$$ which does actually reduce to the usual formula for $m_1\gg m_2$, but is clearly wrong because of the asymmetry about the 2 masses. Moreover, I cant find any source on any such formula. What am is the error in my reasoning here?

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  • $\begingroup$ Work done will change because now the the mass(which you are moving) will have it's own field therefore now you will have to do less work because bodies will attract themselves, Does it help ? $\endgroup$
    – Bhavay
    Jul 8, 2020 at 15:34

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The answer is yes the formula $$U(r) = -G\frac{m_1 m_2}{r}$$ is general where $r$ is the distance between the two masses. In Newtonian gravity the axiom is that the gravitational force felt by body 1 from body 2 is $${\bf F}_{12} = G \frac{m_1 m_2\, {\bf r}}{r^3} = G \frac{m_1 m_2\, ({\bf r_2 - r_1})}{|{\bf r_2}-{\bf r_1}|^3}$$ from that and the definition of potential, $$U({\bf r}_0) = \int_\infty^{{\bf r}_0} {\bf F}\cdot {\rm d}{\bf r},$$ the formula is derived (easiest way is to put one of the masses at the origin, then this integral is trivial). You should think about the potential energy as being assigned to a certain configuration of masses. You can build the configuration by steps. Step 1 there is no mass in the region under consideration so you can bring $m_1$ from infinity to the region of interest without spending any energy, then in step 2 keeping $m_1$ where it, is you bring $m_2$ to its final distance $r$. You should think about this system as being frozen in time, to which we assign the given gravitational potential, so it is a label essentially to a instantaneous configuration, this being a consequence of Newton's version of gravity acting immediately in all of space, but this point will carry us off-topic.

Another question and a completely different matter is, how does this system evolve. If you "release" this system you must now study its dynamics and start making use of Newton's laws and so on to determine equations of motion, trajectories and so on.

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  • $\begingroup$ im sorry if Im being too stupid, but what are the two forms in which you have written the law of gravitation and how are they different from the first formula. Secondly, could you possibly guide me on how t derive this formula without the assumption that one mass stays stationary while the other is brought from infinity? I agree that the field and thus potential can vary over time, but I still dont understand why it is reasonable to equate the change in potential energy to the negative of the work done on just one body, in a 2 body system. $\endgroup$ Jul 8, 2020 at 11:52
  • $\begingroup$ The first formula is gravitational potential, the second set of formulas is gravitational force. About the second point I hope it is clear from the answer, that the potential belongs to the configuration, not to one of the bodies. The configuration is an instant in time if you will, so things don't move. You can also think of the potential as the energy stored in binding the two objects. $\endgroup$
    – ohneVal
    Jul 8, 2020 at 12:02
  • $\begingroup$ oh, im sorry. Didnt realise the first one was potential energy. I meant that the 2 formulas for gravitation look so different than what I have seen on books. Secondly, yes, I do understand the association with the configuration. That still doesnt change how you actually calculate the formula for the potential energy, does it? it is equal to negative of the work done in changing the configuration of the system from when r=infinity to r=R, if im not wrong. What I mean by keeping one mass stationary is that the work done is only computed for one mass, the other dismissed as Fx0displacement $\endgroup$ Jul 8, 2020 at 12:21
  • $\begingroup$ Time is not playing any role in studying the situation of the potential because the force does not depend on time, just on position, therefore at a FIXED time there is no dynamics no velocities no movement. The "bringing" the mass from infinity is just a pedagogical way of picturing how the potential energy can be obtained it will give you the same if you start with mass 2 and they are not "physically" moved towards once another, you are just adding the total work that would be required to set both masses at a given distance. I suggest you first try reviewing other cases of potential energy... $\endgroup$
    – ohneVal
    Jul 8, 2020 at 14:03
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Suppose the (spherically symmetric) bodies are identical and initially separated by distance $R$. We will now take both bodies to infinity synchronously (that is keeping the midpoint in one place – relative to the fixed stars!) If we measure distance $r$ from the centre of mass of the system, that is midway between the centres of the bodies, then the work done on each body taking it from $r=\frac R2$ to infinity is $$\int_{R/2}^\infty\frac{GM_1M_2}{(2r)^2}dr=\frac 14\int_{R/2}^\infty\frac{GM_1M_2}{r^2}dr=\frac 12 \frac{GM_1M_2}{R} $$ So the total work done is $$\frac{GM_1M_2}{R}$$ This is the gain in PE of the system, so its PE in the original configuration was, relative to the PE at infinite separation, $$-\frac{GM_1M_2}{R}$$ With a little more thought we can adapt this treatment to bodies of unequal, but not necessarily hugely unequal, mass.

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  • $\begingroup$ Thankyou for this. I think I understand the mathematical derivation. Could you, perhaps, intuitively explain whats going on here? Why is the work done obtained exactly same if I consider one mass at rest and completely ignore work done on it? $\endgroup$ Jul 8, 2020 at 15:28
  • $\begingroup$ (a) "[...] if I consider one mass at rest and completely ignore work done on it [...]" There won't of course be any work done on it if it is held at rest! (b) The mutual PE of two bodes depends only on their separation, so that's why we have to do the same total amount of work to change the separation from one value to another whether we do work on both bodies or just one body. (c) It occurs to me that your difficulty is as much to do with how we actually bring about the change in separation for bodies of comparable mass as with work and energy. $\endgroup$ Jul 8, 2020 at 15:46
  • $\begingroup$ Considering something at rest, and actually keeping it at rest by exerting forces is different; I am not “holding” it at rest. But you’re right, I am in the process of figuring these concepts out $\endgroup$ Jul 8, 2020 at 16:15

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